$f(x) = \dfrac{ \sqrt{ x - 4 } }{ x^2 + 9 x + 18 }$ What is the domain of the real-valued function $f(x)$ ?
Answer: $f(x) = \dfrac{ \sqrt{ x - 4 } }{ x^2 + 9 x + 18 } = \dfrac{ \sqrt{ x - 4 } }{ ( x + 3 )( x + 6 ) }$ First, we need to consider that $f(x)$ is undefined anywhere where the radical is undefined, so the radicand (the expression under the radical) cannot be less than zero. So $x - 4 \geq 0$ , which means $x \geq 4$ Next, we also need to consider that $f(x)$ is undefined anywhere where the denominator is zero. So $x \neq -3$ and $x \neq -6$ However, these last two restrictions are irrelevant since $4 > -3$ and $4 > -6$ and so $x \geq 4$ will ensure that $x \neq -3$ and $x \neq -6$ Combining these restrictions, then, leaves us with simply $x \geq 4$ Expressing this mathematically, the domain is $\{ \, x \in \RR \mid x \geq4\, \}$.